# How to solve the inequality $\frac {5x+1}{4x-1}\geq1$

\begin{eqnarray} \\\frac {5x+1}{4x-1}\geq1\\ \end{eqnarray}

I have tried the following method but it is wrong. Why?

\begin{eqnarray} \\\frac {5x+1}{4x-1}&\geq&1\\ \\5x+1&\geq& 4x-1\\ \\x &\geq& -2 \end{eqnarray}

• What happens if $x = \frac14$, and what if $x < \frac14$? Oct 16 '13 at 11:39

As Carlos Eugenio Thompson Pinzón has commented, we need $\displaystyle4x-1\ne0$

Method $1:$ If $\displaystyle4x-1\ne0, (4x-1)^2>0$

Multiplying either sides of $\displaystyle\frac{5x+1}{4x-1}\ge1$ by $(4x-1)^2$

we get $\displaystyle(5x+1)(4x-1)\ge(4x-1)^2$

$\displaystyle\implies (4x-1)\{5x+1-(4x-1)\}\ge0$

$\displaystyle\implies \left(x-\frac14\right)\{x-(-2)\}\ge0$

We know if $(x-a)(x-b)\ge0$ where $a<b$ either $x\le a$ or $x\ge b$

Method $2:$

As we know $a\ge b\implies \begin{cases} ac\ge bc &\mbox{if } c> 0 \\ ac\le bc &\mbox{if } c< 0 \end{cases}$

If $\displaystyle 4x-1>0\iff x>\frac14, \frac{5x+1}{4x-1}\ge1\implies 5x+1\ge 4x-1\iff x\ge-2$

$\displaystyle\implies x>\frac14$ is one of the solutions

If $4x-1<0\iff x<\frac14, \frac{5x+1}{4x-1}\ge1\implies 5x+1\le 4x-1\iff x\le-2$

$\displaystyle\implies x\le -2$ is the other solution

• $4x-1$ must be greater then 0, it cann't be 0 Oct 16 '13 at 11:50
• The second row is not equivalent to the first one, I think Oct 16 '13 at 11:58
• So you mean the answers are $x\ge \frac{1}{4}$ or $x\le -2$, right? Oct 16 '13 at 11:59
• Should be $x > 1/4$ or $x \leq -2$: $(-\infty,-2] \cup (\frac{1}{4};+\infty)$ Oct 16 '13 at 12:00
• @CasperLi Yes, but as you do, you have to be extra careful about where $(4x - 1)^2 = 0$. Since the original inequality isn't valid there, neither can the solution be. Oct 16 '13 at 12:38

Hint: $\frac{5x+1}{4x-1} \geq 1$ is equivalent to $\frac{5x+1}{4x-1} -1 \geq 0$ and $\frac{5x+1-(4x-1)}{4x-1} \geq 0$

• Edit that weird "$\,-+1\,$" you wrote in the three numerators. +1 Oct 16 '13 at 11:42
• @DonAntonio: Thanks, I fixed it. Oct 16 '13 at 11:47

Recall that multiplication/divison by negative values in inequalities reverses the sign. We do not know if "4x-1" is a non-negative number such that the sign remains constant when multiplying across - how can you ensure so?

• @DonAntonio I quote from the OP "I have tried the following method but it is wrong. Why?". Aean has given a perfectly correct answer to that question. Oct 16 '13 at 11:49
• Indeed he has, didn't notice that. I'm upvoting this answer and erasing my past comment. Oct 16 '13 at 12:10
• Thank you all! I understand now =) Oct 16 '13 at 12:38

$$\frac {5x+1}{4x-1}\geq1$$ $$\frac {5x+1}{4x-1}-1\geq 0$$

$$\frac {x+2}{4x-1}\geq 0$$ $$x\in(-\infty,-2]\cup(1/4,\infty)$$

• Should be $(1/4; +\infty)$ instead of $[1/4;+\infty)$, I think Oct 16 '13 at 12:22
• user35603, why did you say [1/4, +∞) is not correct? Oct 16 '13 at 12:33
• @CasperLi: since the initial inequality is not defined at $x=1/4$ Oct 16 '13 at 12:49

You easily see that equality holds iff $x=-2$. Then remember that $f(x)=(5x+1)/(4x-1)$ is continuous and not defined at $x=1/4$.

$$\frac{5x+1}{4x-1}\geq 1\Leftrightarrow \frac{5x+1}{4x-1}-1\geq 0\Leftrightarrow\frac{5x+1-(4x-1)}{4x-1}\geq 0 \Leftrightarrow \frac{5x+1-4x+1}{4x-1}\geq 0$$ $$\Leftrightarrow \frac{x+2}{4x-1}\geq 0$$ Now $x+2=0 \Rightarrow x=-2$, and $4x-1=0\Rightarrow x=\frac{1}{4}$

For $x\in (-\infty, -2]$, and $x\in(\frac{1}{4}, +\infty)$, $\frac{x+2}{4x-1}\geq 0$